Determine k, so that k2 + 4k + 8, 2k2 + 3k + 6 and 3k2 + 4k + 4 are three consecutive terms of an AP.
Let a1 = k2 + 4k + 8
a2 = 2k2 + 3k + 6
a3 = 3k2 + 4k + 4
Three terms will be in an AP if
a2 - a1 = a3 - a2
2k2 + 3k + 6 - (k2 + 4k + 8) = 3k2 + 4k + 4 - (2k2 + 3k + 6)
k2 - k - 2 = k2 + k - 2
2k = 0
k = 0